Quiz: Electrical Engineering
Exam: AE/JE
Topic: Miscellaneous
Each question carries 1 mark.
Negative marking: 1/4 mark
Time: 10 Minute
Q1. In squirrel cage induction motor, the rotor conductors are
(a) open circuited
(b) short circuited via end rings
(c) short circuit via external reactance
(d) short circuit via external resistance
Q2. If the frequency of input voltage of a transformer is increased keeping the magnitude of the voltage unchanged, then
(a) both hysteresis loss and eddy current loss in core will increase.
(b) hysteresis loss will increase but eddy current loss will decreases
(c) hysteresis loss will decrease but eddy current loss will remain unchanged
(d) hysteresis loss will decrease but eddy current loss will increase
Q3. A tank circuit consists of
(a) an inductor and a capacitor connected in series
(b) an inductor and a capacitor connected in parallel
(c) a pure inductance and a pure capacitance connected in series
(d) a pure inductance and a pure capacitance connected in parallel
Q5. The average value of the wave V = 110 + 175 sin (314t – 25°) volts is
(a) 110 V
(b) 175 V
(c) 165.57 V
(d) 206.7 V
Q6. Three inductors each of 60 mH are connected in delta. The value of inductance of each arm of the equivalent star connection is
(a) 10 mH
(b) 15 mH
(c) 20 mH
(d) 30 mH
Q7. The magnetic field energy in an inductor changes from maximum value to minimum value in 5msec when connected to an AC source. The frequency of the source in Hz is
(a) 500
(b) 200
(c) 50
(d) 20
Q8. An amplifier has a gain of 20 without feedback. If 10% of the output voltage is fed back by means of resistance negative feedback circuit, the overall gain would be
(a) 19.80
(b) 16.55
(c) 10.85
(d) 6.67
Q9. Phase reversed of 180⁰ as compared to input occurs in the output of
(a) CE amplifier
(b) CB amplifier
(c) CC amplifier
(d) All of the above
Q10. The impulse response of a second order under-damped system starting from rest is given by:
C(t)=12.5 e^(-6t) sin8t;t≥0
What is the value of natural frequency and damping factor of the system, respectively?
(a) 10 units and 0.6
(b) 10 units and 0.8
(c) 8 units and 0.6
(d) 8 units and 0.8
solutions
S1. Ans.(b)
Sol. in squirrel – cage induction motor, the rotor conductors are short circuit via end rings.
The rotor conductors are not wires, but it consists of heavy bars of copper, aluminum, or an alloy.
If the rotor and the stator conductors are parallel to each other, there is a strong possibility of the magnetic locking between the rotor and the stator. Therefore, the rotor slots are skewed.S2. Ans.(c)
Sol. if the frequency of input voltage of a transformer is increased keeping the magnitude of the voltage unchanged, the hysteresis loss will decrease but eddy current loss will remain unchanged.
In the given question, voltage kept constant and frequency is varied.
V/f ratio is not constant.
Hysteresis losses, P_h ∝ V^1.6 f^(-0.6)
As frequency is increased, the hysteresis losses will decrease.
Eddy current losses, We ∝ V^2
As eddy current losses are independent of frequency, the eddy current losses remain the same.
S3. Ans.(d)
Sol. Tank circuit
S4. Ans.(c)
Sol. Speed torque characteristics of dc series motor:
S5. Ans.(a)
Sol. V = 110 + 175 sin (314 t – 25°)
V_avg=110+0=110V
S6. Ans.(c)
Sol.
L_eqv=L/3=60/3×10^(-3)=20mH
S7. Ans.(c)
Sol. The magnetic energy in an inductor is given by,
E=1/2 LI^2
The magnetic energy will be maximum when the current will reach its peak value Io and it will be minimum when the current will become zero.
From above graph,
T/4=5 ms
⇒T=20 ms
∵f=1/T=1/(20×10^(-3) )=1000/20
=50 Hz
S8. Ans.(d)
Sol. Given that, A=20 and β=10%=0.1
∵feedback is Negative.
∴A_v=A/(1+Aβ)=20/(1+20×0.1 )=20/3=6.67
S9. Ans.(a)
Sol. Phase reversed of 180⁰ as compared to input occurs in the output of CE amplifier.
